Second Order Linear Partial Differential Equations
Part II
Fourier series; Euler-Fourier formulas; Fourier Convergence Theorem;
Even and odd functions; Cosine and Sine Series Extensions; Particular
solution of the heat conduction equation
Fourier Series
Suppose f is a periodic function with a period T = 2L. Then the Fourier
series representation of f is a trigonometric series (that is, it is an infinite
series consists of sine and cosine terms) of the form
a0 ∞ 
nπ x
nπ x 
f ( x ) = + ∑  a n cos
+ bn sin

L
L 
2 n =1 
Where the coefficients are given by the Euler-Fourier formulas:
1
mπ x
am = ∫ f ( x) cos
dx , m = 0, 1, 2, 3, …
L −L
L
L
1
nπ x
bn = ∫ f ( x) sin
dx ,
L −L
L
L
n = 1, 2, 3, …
The coefficients a’s are called the Fourier cosine coefficients (including a0,
the constant term, which is in reality the 0-th cosine term), and b’s are called
the Fourier sine coefficients.
© 2008, 2012 Zachary S Tseng
E-2 - 1
Note 1: Thus, every periodic function can be decomposed into a sum of one
or more cosine and/or sine terms of selected frequencies determined solely
by that of the original function. Conversely, by superimposing cosines and/
or sines of a certain selected set of frequencies we can reconstruct any
periodic function.
Note 2: If f is piecewise continuous, then the definite integrals in the EulerFourier formulas always exist (i.e. even in the cases where they are improper
integrals, the integrals will converge). On the other hand, f needs not to be
piecewise continuous to have a Fourier series. It just needs to be periodic.
However, if f is not piecewise continuous, then there is no guarantee that we
could find its Fourier coefficients, because some of the integrals used to
compute them could be improper integrals which are divergent.
Note 3: Even though that the “=” sign is usually used to equate a periodic
function and its Fourier series, we need to be a little careful. The function f
and its Fourier series “representation” are only equal to each other if, and
whenever, f is continuous. Hence, if f is continuous for −∞ < x < ∞, then f is
exactly equal to its Fourier series; but if f is piecewise continuous, then it
disagrees with its Fourier series at every discontinuity. (See the Fourier
Convergence Theorem below for what happens to the Fourier series at a
discontinuity of f .)
Note 4: Recall that a function f is said to be periodic if there exists a positive
number T, such that f (x + T ) = f (x), for all x in its domain. In such a case
the number T is called a period of f. A period is not unique, since if f (x + T )
= f (x), then f (x + 2T ) = f (x) and f (x + 3T ) = f (x) and so on. That is, every
integer-multiple of a period is again another period. The smallest such T is
called the fundamental period of the given function f. A special case is the
constant functions. Every constant function is clearly a periodic function,
with an arbitrary period. It, however, has no fundamental period, because its
period can be an arbitrarily small real number. The Fourier series
representation defined above is unique for each function with a fixed period
T = 2L. However, since a periodic function has infinitely many (nonfundamental) periods, it can have many different Fourier series by using
different values of L in the definition above. The difference, however, is
really in a technical sense. After simplification they would look the same.
© 2008, 2012 Zachary S Tseng
E-2 - 2
Therefore, technically at least, a Fourier series of a periodic function
depends both on the function as well as its chosen period.
Note 5: The definite integrals in the Euler-Fourier formulas can be found be
integrating over any interval of length 2L. However, from −L to L is the
convention, and is often the most convenient interval to use.
Note 6: Since the Fourier coefficients are calculated by definite integrals,
which are insensitive to the value of the function at finitely many points.
Consequently, piecewise continuous functions of the same period that differ
from each other at finitely many points (notably, at isolated discontinuities)
per period will have the same Fourier series.
Note 7: The constant term in the Fourier series, which has expression
L
L
a0 1 1
1
= ⋅ ∫ f ( x) cos(0) dx =
f ( x) dx ,
2 2 L −L
2 L −∫L
is just the average or mean value of f (x) on the interval [−L, L]. Since f is
periodic, this average value is the same for every period of f. Therefore, the
constant term in a Fourier series represents the average value of the function
f over its entire domain.
© 2008, 2012 Zachary S Tseng
E-2 - 3
Example: Find a Fourier series for f (x) = x, −2 < x < 2, f (x + 4) = f (x).
First note that T = 2L = 4, hence L = 2.
The constant term is one half of:
1
mπ x
1
a0 = ∫ f ( x) cos
dx =
L −L
L
2
L
2
1 x2
∫−2 x dx = 2 2
2
=
−2
1
( 2 − 2) = 0
2
The rest of the cosine coefficients, for n = 1, 2, 3, …, are
1
nπ x
1
an = ∫ f ( x) cos
dx =
L −L
L
2
L
1  2 x
nπx
2
=
sin
−
2  nπ
2 − 2 nπ
2
2
∫ x cos
−2
nπx 
∫−2sin 2 dx 

2
1  2x
nπx
4
nπx
= 
sin
+ 2 2 cos
2  nπ
2
nπ
2 −2
2
=
nπx
dx
2




1 
4
4
 

  0 + 2 2 cos( nπ )  −  0 + 2 2 cos( − nπ )   = 0
2 
nπ
nπ
 

Hence, there is no nonzero cosine coefficient for this function. That is,
its Fourier series contains no cosine terms at all. (We shall see the
significance of this fact a little later.)
© 2008, 2012 Zachary S Tseng
E-2 - 4
The sine coefficients, for n = 1, 2, 3, …, are
1
nπ x
1
bn = ∫ f ( x) sin
dx =
L −L
L
2
L
2
∫ x sin
−2
nπ x
dx
2
2
2
1  − 2 x
nπ x
−2
nπ x 
=
cos
−
cos
dx

2  nπ
2 −2 nπ −∫2
2

2
nπ x
nπ x
1  − 2x
4
= 
cos
+ 2 2 sin
nπ
2  nπ
2
2 −2




=
1  − 4
  4

cos( nπ ) − 0  − 
cos( − nπ ) − 0  
 
2   nπ
  nπ

=
−2
(cos(nπ ) + cos(nπ ) ) = − 4 cos( nπ )
nπ
nπ
 4
 nπ , n = odd
=
−4
 , n = even
 nπ
Therefore, f ( x ) =
© 2008, 2012 Zachary S Tseng
(−1) n +1 4
=
nπ .
nπ x
(−1) n + 1
sin
.
∑
2
π n =1 n
4
∞
E-2 - 5
Figure: the graph of the partial sum of the first 30 terms of the
Fourier series
nπ x
(−1) n + 1
f ( x) = ∑
sin
.
2
π n =1 n
4
∞
Compare it against the graph of the actual function the series
represents the function f (x) = x, −2 < x < 2, f (x + 4) = f (x), seen
earlier.
© 2008, 2012 Zachary S Tseng
E-2 - 6
Example: Find a Fourier series for f (x) = x, 0 < x < 4, f (x + 4) = f (x). How
will it be different from the series above?
4
4
1
1 x2
a0 = ∫ x dx =
20
2 2
=
0
1
(8 − 0) = 4
2
For n = 1, 2, 3, … :
4
1
nπx
an = ∫ x cos
dx
20
2
1  2x
nπx
4
nπx
= 
sin
+ 2 2 cos
2  nπ
2
nπ
2 0
4
=




1 
4
4
 

  0 + 2 2 cos( 2nπ )  −  0 + 2 2 cos(0)   = 0
2 
nπ
nπ
 

4
1
nπx
bn = ∫ x sin
dx
20
2
1  − 2x
nπx
4
nπx
= 
cos
+ 2 2 sin
2  nπ
2
nπ
2 0
4
=




 −4
1  − 8

cos( 2nπ ) − 0  − (0 − 0 ) =
 
2   nπ

 nπ
Consequently,
a0 ∞ 
nπ x
nπ x 
−4 ∞ 1
nπx
f ( x) = + ∑  an cos
+ bn sin
sin
 = 2+
∑
2 n =1 
L
L 
2 .
π n =1 n
© 2008, 2012 Zachary S Tseng
E-2 - 7
Example: Find a Fourier series for f (x) = | x |, −2 < x < 2, f (x + 4) = f (x).
Answer : f ( x) = 1 −
8
π
2
∞
1
∑ (2n − 1)
n =1
2
cos
(2n − 1)π x
2
Example: Find a Fourier series for
 − 2,
f ( x) = 
 2,
−1 ≤ x < 0
,
0 ≤ x <1
Answer : f ( x) =
© 2008, 2012 Zachary S Tseng
8
∞
f (x + 2) = f (x).
1
∑ (2n − 1) sin((2n − 1)π x)
π
n =1
E-2 - 8
Comment: Just because a Fourier series could have infinitely many (nonzero)
terms does not mean that it will always have that many terms. If a periodic
function f can be expressed by finitely many terms normally found in a
Fourier series, then the expression must be the Fourier series of f. (This is
analogous to the fact that the Maclaurin series of any polynomial function is
just the polynomial itself, which is a sum of finitely many powers of x.)
Example: The Fourier series (period 2π) representing f (x) = 5 + cos(4x) −
sin(5x) is just f (x) = 5 + cos(4x) − sin(5x).
Example: The Fourier series (period 2π) representing f (x) = 6 cos(x) sin(x) is
not exactly itself as given, since the product cos(x) sin(x) is not a term in a
Fourier series representation. However, we can use the double-angle
formula of sine to obtain the result: 6 cos(x) sin(x) = 3 sin(2x).
Consequently, the Fourier series is f (x) = 3 sin(2x).
© 2008, 2012 Zachary S Tseng
E-2 - 9
The Fourier Convergence Theorem
Here is a theorem that states a sufficient condition for the convergence of a
given Fourier series. It also tells us to what value does the Fourier series
converge to at each point on the real line.
Theorem: Suppose f and f ′ are piecewise continuous on the interval
−L ≤ x ≤ L. Further, suppose that f is defined elsewhere so that it is periodic
with period 2L. Then f has a Fourier series as stated previously whose
coefficients are given by the Euler-Fourier formulas. The Fourier series
converge to f (x) at all points where f is continuous, and to
 lim f ( x) + lim f ( x) / 2
 x → c −

x → c+
at every point c where f is discontinuous.
Comment: As seen before, the fact that f is piecewise continuous guarantees
that the Fourier coefficients can be found. The condition that f ′ is also
piecewise continuous is a sufficient condition to guarantee that the series
thusly found will be convergent everywhere on the real line. As well, recall
that, suppose f is continuous at c, then by definition f (c) equals both onesided limits of f (x) as x approaches c. Therefore, the second part of the
theorem could be even more succinctly stated as that the Fourier series
representing f will always converge to
 lim f ( x) + lim f ( x) / 2
 x → c −

x → c+
at every point c (and not just at discontinuities of f ).
A consequence of this theorem is that the Fourier series of f will “fill in” any
removable discontinuity the original function might have. A Fourier series
will not have any removable-type discontinuity.
© 2008, 2012 Zachary S Tseng
E-2 - 10
Example: Let us revisit the earlier calculation of the Fourier series
representing f (x) = x,
−2 < x < 2,
f (x + 4) = f (x).
The Fourier series, as we have found, is
nπ x
(−1) n + 1
f ( x) = ∑
sin
.
π n =1 n
2
4
∞
The following figures are the graphs of various finite n-th partial sums of the
series above.
n=3
n = 10
© 2008, 2012 Zachary S Tseng
E-2 - 11
n = 20
n = 30
n = 50
© 2008, 2012 Zachary S Tseng
E-2 - 12
Note that superimposed sinusoidal curves take on the general shape of the
piecewise continuous periodic function f (x) almost immediately. As well,
for the parts of the curve where f (x) is continuous (where the Fourier
Convergence Theorem predicts a perfect match) the composite curve of the
Fourier series converges rapidly to that of f (x), as predicted. The
convergence is not as rapidly near the jump discontinuities. Indeed, for all
but the lowest partial sums of the Fourier series, the curve seems to
“overshoot” that of f (x) near each jump discontinuity by a noticeable margin.
Further more, this discrepancy does not fade away for any finitely larger n.
That is, the convergence of a Fourier series, while predictable, is not uniform.
(That is a small price we pay for approximating a piecewise continuous
periodic function by sinusoidal curves. It can be done, but the Fourier series
does not converge uniformly to the actual function.)
This behavior is known as the Gibbs Phenomenon. It further states that the
partial sums of a Fourier series will overshoot a jump discontinuity by an
amount approximately equal to 9% of the jump. That is, near each jump
discontinuity, the overshoot amounts to about
0.09 lim+ f ( x) − lim− f ( x) ,
x→c
x→c
for large n. Further, this overshoot does not go away for any finitely large n.
© 2008, 2012 Zachary S Tseng
E-2 - 13
Question: Sketch the graph of the Fourier series of
f (x) = x,
−2 < x < 2, f (x + 4) = f (x).
We have seen a few graphs of its partial sums. But what will the graph of
the actual Fourier series look like?
Example: Sketch the graph of the Fourier series of
f (x) = | x |, −2 < x < 2, f (x + 4) = f (x).
Example: Sketch the graph of the Fourier series of
 − 2,
f ( x) = 
 2,
© 2008, 2012 Zachary S Tseng
−1 ≤ x < 0
,
0 ≤ x <1
f (x + 2) = f (x).
E-2 - 14
Even and Odd Functions
Recall that an even function is any function f such that
f (−x) = f (x),
for all x in its domain.
Examples: cos(x), sec(x), any constant function, x2, x4, x6, … , x −2, x −4, …
An odd function is any function f such that
f (−x) = −f (x),
for all x in its domain.
Examples: sin(x), tan(x), csc(x), cot(x), x, x3, x5, … , x −1, x −3, …
Most functions, however, are neither even nor odd. There is one function
that is both even and odd. (What is it?)
Arithmetic Combinations of Even and Odd Functions
The table below summaries the result of performing the common arithmetic
operations on a pair of even and/or odd functions:
+ / −
× / ÷
Even and Even
Even
Even
Odd and Odd
Odd
Even
Even and Odd
Neither
Odd
The result above can be extended to arbitrarily many terms. For example, a
sum of three or more even functions will again be even. (Care needs to be
taken in the cases where 3 or more odd functions forming a product/quotient.
For example, a product of 3 odd functions will be odd, but a product of 4
odd functions is even.)
© 2008, 2012 Zachary S Tseng
E-2 - 15
Calculus Properties of Even and Odd Functions
Suppose f is an even function, continuous on −L ≤ x ≤ L, then
L
∫
−L
L
f ( x ) dx = 2 ∫ f ( x) dx .
0
Suppose f is an odd function, continuous on −L ≤ x ≤ L, then
L
∫
f ( x ) dx = 0 .
−L
© 2008, 2012 Zachary S Tseng
E-2 - 16
The Fourier Cosine Series
Suppose f is an even periodic function of period 2L, then its Fourier series
contains only cosine (include, possibly, the constant term) terms. It will not
have any sine term. That is, its Fourier series is of the form
a0 ∞
nπ x
f ( x) = + ∑ an cos
2 n =1
L .
Conversely, any periodic function whose Fourier series has the form of a
cosine series as shown must be an even periodic function. Computationally,
this means that the Fourier coefficients of an even periodic function are
given by
1
mπ x
2
mπ x
am = ∫ f ( x) cos
dx = ∫ f ( x) cos
dx ,
L −L
L
L0
L
L
L
m = 0, 1, 2, 3, …
bn = 0,
n = 1, 2, 3, …
Notice that the integrand in the definite integral used to find the cosine
coefficients a’s is an even function (it is a product of two even functions, f (x)
and cos x). Therefore, we can use the symmetric property of even functions
to simplify the integral.
© 2008, 2012 Zachary S Tseng
E-2 - 17
The Fourier Sine Series
If f is an odd periodic function of period 2L, then its Fourier series contains
only sine terms. It will not have any cosine term. That is, its Fourier series
is of the form
∞
f ( x) = ∑ bn sin
n =1
nπ x
L
Conversely, any periodic function whose Fourier series has the form of a
sine series as shown must be an odd periodic function. Therefore, the
Fourier coefficients of an odd periodic function are given by
am = 0,
2
bn =
L
m = 0, 1, 2, 3, …
L
∫
f ( x) sin
0
nπ x
dx ,
L
n = 1, 2, 3, …
Example: We have calculated earlier that the function f (x) = x, −2 < x < 2,
f (x + 4) = f (x), has as its Fourier series consists of purely sine terms:
nπ x
4 ∞ (−1) n + 1
f ( x) = ∑
sin
.
π n =1 n
2
We now see that this sine series signifies that the function is odd periodic.
It is perhaps not very obvious, but the integrand in the integral for the
Fourier sine coefficients is another even function. It is a product of two odd
functions, f (x) and sin x, which makes it even. Therefore, we can again take
advantage of the symmetric property of even functions to simplify the
integral.
© 2008, 2012 Zachary S Tseng
E-2 - 18
The Cosine and Sine Series Extensions
If f and f ′ are piecewise continuous functions defined on the interval
0 ≤ t ≤ L, then f can be extended into an even periodic function, F, of period
2L, such that f (x) = F(x) on the interval [0, L], and whose Fourier series is,
therefore, a cosine series. Similarly, f can be extended into an odd periodic
function of period 2L, such that f (x) = F(x) on the interval (0, L), and whose
Fourier series is, therefore, a sine series. The process that such extensions
are obtained is often called cosine /sine series half-range expansions.
Here is an outline of how this can be done. Start with a function that is
defined only on an interval of finite length, from 0 to L. First expand the
function to be defined on the interval from −L to L such that the function is
an even or an odd function as required. Then define the function to be
periodic with a period of T = 2L by requiring F(x + 2L) = F(x). This process
is actually much easier than it sounds. Mathematically, the process can be
achieved rather simply, as described below.
Even (cosine series) extension of f (x)
Given f (x) defined on [0, L]. Its even extension of period 2L is:
Where
 f ( x), 0 ≤ x ≤ L
F ( x) = 
,
 f (− x), − L < x < 0
F(x + 2L) = F(x).
a0 ∞
nπ x
F ( x) = + ∑ an cos
,
L
2 n =1
such that
2
am =
L
bn = 0,
© 2008, 2012 Zachary S Tseng
L
∫
0
f ( x) cos
mπ x
dx ,
L
m = 0, 1, 2, 3, …
n = 1, 2, 3, …
E-2 - 19
Odd (sine series) extension of f (x)
Given f (x) defined on (0, L). Its odd extension of period 2L is:
0< x<L
 f ( x),

F ( x) =  0 ,
x = 0, L ,
− f (− x), − L < x < 0

∞
F ( x ) = ∑ bn sin
Where
n =1
nπ x
,
L
such that
am = 0,
2
bn =
L
m = 0, 1, 2, 3, …
L
∫
f ( x) sin
0
Example: Let f (x) = x,
extensions of period 4.
Answers:
F(x + 2L) = F(x).
nπ x
dx ,
L
n = 1, 2, 3, …
0 ≤ x < 2. Find its cosine and sine series
1
(2n − 1)π x
cos
∑
π 2 n =1 (2n − 1) 2
2
8
∞
Cosine series:
f ( x) = 1 −
Sine series:
nπ x
(−1) n + 1
f ( x) = ∑
sin
π n =1 n
2
© 2008, 2012 Zachary S Tseng
4
∞
E-2 - 20
Back to the Heat Conduction Problem
Previously, we had found the general solution of the initial-boundary value
problem given by the one-dimensional heat conduction equation modeling a
bar that has both of its ends kept at 0 degree. The general solution is
∞
u ( x, t ) = ∑ C n e
−α 2 n 2π 2 t / L2
n =1
sin
nπ x
L .
Setting t = 0 and applying the initial condition u(x, 0) = f (x), we get
∞
u ( x,0) = ∑ Cn sin
n =1
nπ x
= f ( x) .
L
We now know that the above equation says that the initial condition needs to
be an odd periodic function of period 2L. Since the initial condition could
be an arbitrary function, it usually means that we would need to “force the
issue” and expand it into an odd periodic function of period 2L. That is
∞
f ( x ) = ∑ bn sin
n =1
nπ x
L .
Compare the two expressions, we see that
∞
u ( x,0) = ∑
n =1
∞
nπ x
nπ x
Cn sin
= f ( x) = ∑ bn sin
L
L .
n=1
Therefore, the particular solution is found by setting all the coefficients
Cn = bn, where bn’s are the Fourier sine coefficients of (or the odd periodic
extension of) the initial condition f (x):
2
nπ x
Cn = bn = ∫ f ( x) sin
dx .
L0
L
L
© 2008, 2012 Zachary S Tseng
E-2 - 21
Example: Solve the heat conduction problem
8 uxx = ut
,
0 < x < 5,
u(0, t) = 0, and u(5, t) = 0,
t > 0,
u(x, 0) = 2sin(πx) − 4sin(2πx) + sin(5πx).
2
Since the standard form of the heat conduction equation is α uxx = ut,
2
we see that α = 8; and we also note that L = 5. Therefore, the general
solution is
∞
u ( x, t ) = ∑ Cn e
−α 2 n 2π 2 t / L2
sin
n =1
∞
= ∑ Cn e
−8 n 2π 2 t / 25
n =1
sin
nπ x
L
nπ x
5
The initial condition, f (x), is already an odd periodic function (notice
that it is a Fourier sine series) of the correct period T = 2L = 10.
Therefore, no additional calculation is needed, and all we need to do is
to extract the correct Fourier sine coefficients from f (x). To wit
C5 = b5 = 2,
C10 = b10 = −4,
C25 = b25 = 1,
Cn = bn = 0, for all other n, n ≠ 5, 10, or 25.
Hence,
u ( x, t ) = 2e−8(5
+ e−8( 25
2
)π 2 t / 25
© 2008, 2012 Zachary S Tseng
2
)π 2 t / 25
sin(πx) − 4e−8(10
2
)π 2 t / 25
sin(2πx)
sin(5πx)
E-2 - 22
What will the particular solution be if the initial condition is u(x, 0) = x
instead? That is, solve the following heat conduction problem:
8 uxx = ut
,
0 < x < 5, t > 0,
u(0, t) = 0, and u(5, t) = 0,
u(x, 0) = x.
The general solution is still
∞
u ( x, t ) = ∑ Cn e−8n π
2
2
t / 25
sin
n =1
nπ x
5 .
The initial condition is an odd function, but it is not a periodic
function. Therefore, it needs to be expanded into its odd periodic
extension of period 10 (T = 2L). Its coefficients are, for n = 1, 2, 3, …
2
bn =
L
L
∫
0
5
nπ x
2
nπ x
f ( x ) sin
dx = ∫ x sin
dx
L
50
5
5
5
2  − 5 x
nπ x
−5
nπ x 
=
cos
−
cos
dx

5  nπ
5 0 nπ ∫0
5

2  − 5x
nπ x
25
nπ x
cos
= 
+ 2 2 sin
5  nπ
5
nπ
5
=

2   − 25

cos( nπ ) − 0  − (0 − 0 )
 
5   nπ


=
− 10
cos( nπ )
nπ
 10
 nπ , n = odd
=
− 10

, n = even
 nπ
© 2008, 2012 Zachary S Tseng
5
0




(−1) n+110
=
nπ
E-2 - 23
The resulting sine series is (representing the function f (x) = x, −5 < x
< 5, f (x + 10) = f (x)):
(−1) n +1
nπ x
f ( x) = ∑
sin
.
π n =1 n
5
10
∞
The particular solution can then be found by setting each coefficient,
Cn, to be the corresponding Fourier sine coefficient of the series above,
(−1) n+110
Cn = bn =
. Therefore, the particular solution is
nπ
u( x, t ) =
© 2008, 2012 Zachary S Tseng
10
π
∞
∑
n =1
nπ x
(−1)n +1 −8n 2π 2t / 25
e
sin
n
5 .
E-2 - 24
Exercises E-2.1:
1 – 8 Find the Fourier series representation of each periodic function.
Determine the values to which each series converge to at x = 0.
1. f (x) = 5,
−1 < x < 1,
f (x + 2) = f (x).
2
−π < x < π,
f (x + 2π) = f (x).
2. f (x) = (cos x + sin x) ,
0 ≤ x < 2,
f (x + 2) = f (x).
3. f (x) = 6 − 3x,
2
4. f (x) = x ,
0 ≤ x < π,
f (x + π) = f (x).
2
5. f (x) = x ,
−π < x < π,
f (x + 2π) = f (x).
 0, − 2 < x < 0

6. f ( x) = 4, 0 ≤ x ≤ 1 ,
f (x + 4) = f (x).
 0, 1 < x < 2

7. f (x) = │sin x│,
−π < x < π,
f (x + 2π) = f (x).
8. f (x) = δ(x − c),
−π < x < π, 0 < c < π,
f (x + 2π) = f (x).
9 – 12 Expand each function into its cosine series and sine series
representations of the indicated period. Determine the values to which each
series converge to at x = 0, x =2, and x = −2.
9. f (x) = 3 − x, T = 6.
10. f (x) = e, T = 2π.
 x, 0 ≤ x < 2
11. f (x) = sin x, T = 2π.
12. f ( x) = 
, T = 6.
 2, 2 ≤ x < 3
13. Solve the heat conduction problem
2 uxx = ut
,
0 < x < 9,
t > 0,
u(0, t) = 0, and u(9, t) = 0,
u(x, 0) = 25sin(πx / 3) + 45sin(4πx / 3) − 12sin(3πx).
14. Solve the heat conduction problem of the given initial conditions.
9 uxx = ut
,
0 < x < 12,
u(0, t) = 0, and u(12, t) = 0,
(a)
(b)
(c)
t > 0,
u(x, 0) = 3sin(πx) − sin(7πx / 6) − 6sin(2πx),
u(x, 0) = 4,
u(x, 0) = 4 − x.
© 2008, 2012 Zachary S Tseng
E-2 - 25
Answers E-2.1:
1. f (x) = 5,
f (0) = 5.
2. f (x) = 1 + sin(2x),
f (0) = 1.
∞
6
1
3. f ( x) = 3 + ∑ sin( nπx) ,
f (0) = 3.
π n =1 n
π2
π
1

+ ∑  2 cos(2nx ) − sin( 2nx ) ,
4. f ( x) =
3 n =1  n
n

∞
π2
.
4
∞
1
f (0) = 0.
 nπ   nπx  1 
 nπ
 + 1 − cos
 cos
2   2  n
 2
∑  n sin 
π
n =1

f (0) = 2.
2 4 ∞
1
cos( 2nx ) ,
7. f ( x) = − ∑
π π n =1 (2n − 1)(2n + 1)
8. f ( x) =
2
∞
(−1) n
+ 4 ∑ 2 cos( nx ) ,
5. f ( x) =
3
n =1 n
6. f ( x) = 1 +
f ( 0) =
π2
1 1 ∞
+ ∑ cos( n( x − c )) ,
2π π n =1
   nπx 
  sin 
 ,
   2 
f (0) = 0.
f (0) = 0.
3 ∞
12
 nπx 
cos
,
9. Cosine series: f ( x ) = + ∑
2 2
2 n =1 ( 2n − 1) π
 3 
∞
9
 nπx 
sin 
;
Sine series: f ( x ) = ∑
 3 
n = 1 nπ
The cosine series converges to 3, 1, and 1; the sine series converges to 0, 1,
and −1, respectively, at x = 0, x =2, and x = −2.
10. Cosine series: f (x) = e,
∞
4e
sin( nx) ;
Sine series: f ( x ) = ∑
n =1 ( 2 n − 1)π
The cosine series converges to e at all 3 points. The sine series converges
to 0, e, and − e, respectively, at x = 0, x =2, and x = −2.
2 4 ∞
1
f
(
x
)
=
− ∑
cos(2nx ) ,
11. Cosine series:
π π n =1 ( 2n − 1)(2n + 1)
Sine series: f (x) = sin x;
Both series converge to 0 at x = 0 and to sin(2) at x = 2. At x = −2, the
cosine series converges to sin(2), the sine series to − sin(2).
© 2008, 2012 Zachary S Tseng
E-2 - 26
4 ∞ 6   2nπ    nπx 
 − 1 cos
,
12. Cosine series: f ( x ) = + ∑ 2 2  cos
3 n =1 n π   3    3 
n
∞
 6
 2nπ  4( −1)   nπx 
sin 
−
;
Sine series: f ( x) = ∑  2 2 sin 
nπ   3 
 3 
n =1  n π
Both series converge to 0 at x = 0 and to 2 at x = 2. At x = −2, the cosine
series converges to 2, the sine series to − 2.
2
2
πx
4πx
−2π 2t / 9
sin( ) + 45e −32π t / 9 sin( ) −12e −18π t sin(3πx)
13. u( x, t ) = 25e
3
3
7πx
−9π 2t
− 49π 2 t / 4
−36π 2t
u
(
x
,
t
)
=
3
e
sin(
π
x
)
−
e
sin(
)
−
6
e
sin(2πx) ;
14. (a)
6
16 ∞ 1 −9n 2π 2 t / 144 nπ x
e
sin
(b) u( x, t ) = ∑
.
π n =1 2n − 1
12
© 2008, 2012 Zachary S Tseng
E-2 - 27